Here is the whole problem:

In the triangle $ABC$, it is known that $AC > AB$, and the angle at the vertex $A$ is equal to $\alpha$. On the side $AC$, point $M$ is marked so that $AB=MC$. Point $E$ is the midpoint of the segment $AM$, point $D$ is the midpoint of the segment $BC$. Find the angle $CED$.

The textbook answer to the problem is that $\angle CED = \dfrac{\alpha}{2}$. The tip on how get to it is to mark a point $K$ on the $AB$ such, that $AK = KB$, draw the midline $KD$ of the triangle $ABC$ and then prove that the $KDE$ triangle is isosceles.

At that point I've tried everything(even chapgpt, which, as it turned out, is bad at math), the best I've gotten so far is if I mark a midpoint F on the $CM$, then I get parallelogram $EKDF$ $\bigg(KD = \dfrac{1}{2}AC = EF$, and $KD\parallel AC\bigg)$. And triangles $KDE$ and $FED$ are congruent, but it's not even close to what I need to prove. How would you do that ?

Here is the final figure: enter image description here

Please, take into account that this is a problem from an $\mathbf{8^{th}}$ grade math textbook, the topic is "Midline of a triangle", which means I'm not allowed to use any angle functions or similar features that were not introduced yet in the curicullumn.

EDIT: The textbook's hint seems to be wrong, thanks @Vasili for the solution. The actual triangle that needs to be proven to be isosceles is another one, not the $KDE$.

  • 1
    $\begingroup$ Could you provide more details like what you have studied under the topic mentioned. $\endgroup$
    – KKT
    Commented May 2, 2023 at 14:13
  • $\begingroup$ @KKT, should I list all the material I've gone through ? :) I think I gave sufficient amount of information to the particular problem ? $\endgroup$ Commented May 2, 2023 at 14:31
  • $\begingroup$ The hint “Prove that the $KDE$ triangle is isosceles” seems to be a mistake. The key, as @Vasili shows, is rather that $\triangle NDE$ is isosceles. $\endgroup$ Commented May 2, 2023 at 22:21
  • $\begingroup$ @EdwardPorcella, yeah, that's what I figured when I saw @Vasili solution. $\endgroup$ Commented May 2, 2023 at 22:40

2 Answers 2


Here is how I arrived to the solution but also without completely using the hint.
Draw $BM$ and $NE$. $NE$ is a midline in $\triangle ABM$ so $AKNE$ is a paralellogram, $NE=AK$. $ND$ is a midline in $\triangle BMC$ so $ND=\frac{MC}{2}=AK=NE \implies \triangle END$ is isosceles. enter image description here

  • $\begingroup$ I have no time to review it completely right now, but without BM I understand the proof. I'll check it up later. Thanks for the answer. $\endgroup$ Commented May 2, 2023 at 15:26
  • $\begingroup$ $BM$ is needed to show that $ND=\frac{MC}{2}$ $\endgroup$
    – Vasili
    Commented May 2, 2023 at 15:29
  • $\begingroup$ I just checked your solution, it's correct, thank you very much! I guess the textbook's hint is just wrong But I'd like to add that BM seems to be redundant here. Simply mark point $N$ so that $KN = AE $ => $ND = (KD=EF) - (KN=EM) = MF$. And yeah, from then on we simply prove that $\triangle END$ is isosceles. Thanks again! $\endgroup$ Commented May 2, 2023 at 23:02

enter image description here

As $\triangle BKD \sim \triangle BAC$, $|KD|=L_1+L_3$.

Translating D to the left by $L_3$, and moving E correspondingly (i.e. to A), maintains $\angle DEF$ and creates a parallelogram with all sides equal to $L_1$, of which the translated line $D'E'$ is a diagonal bisector.

  • $\begingroup$ what is the "right line" and the "translated line"? $\endgroup$
    – D S
    Commented May 2, 2023 at 13:02
  • $\begingroup$ Thanks for your answer, but there were no similarity of triangles yet in the curicullumn. I don't really understand what "L1", "L3", "moving E", "translated line" mean. I provided a hint from the textbook, would be nice to see a solution in accordance to that hint. $\endgroup$ Commented May 2, 2023 at 13:06
  • $\begingroup$ Brilliant answer(+1) $\endgroup$
    – D S
    Commented May 2, 2023 at 13:21
  • $\begingroup$ @DS; I haven't. The orange line completes the parallelogram from $AKD'$. $|KD'|=L_1$, as does $|AK|$. $\endgroup$
    – JMP
    Commented May 2, 2023 at 13:41
  • 1
    $\begingroup$ I think we can do this without translations and similarity, but employing the same idea. But it still does not use the hint. $\endgroup$
    – D S
    Commented May 2, 2023 at 13:58

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